Biology IIH
Cell Size Laboratory Investigation
Student Guide
Background
Cells come in a variety of sizes. Some red blood cells are only 8 µm in diameter.
Nerve cells can reach lengths up to 1 meter. Most living cells, however, are between
2 and 200 µm in diameter. Why can’t organisms be just one giant cell?
Diffusion limits cell size: cells require a constant supply of glucose and oxygen to
carry out cellular respiration and to produce large amounts of energy. These
substances, and waste products such as carbon dioxide, move through the
cytoplasm by diffusion. It has been estimated that it takes a molecule of oxygen only
a fraction of a second to diffuse through the cytoplasm from the plasma membrane to
the center of a typical cell that is 20 µm. What would happen if the cell got bigger?
Although diffusion is efficient over short distances, it becomes slow and inefficient as
the distance becomes larger. Hypothetically, a cell with a diameter of 20 cm would
have to wait months before receiving molecules that enter the cell. Because of this
time restriction the cell/organisms would die if they were one large cell.
A cell’s DNA limits size: The second reason why cells are small is because they
usually only contain one nucleus. If a cell doesn’t have enough DNA to program its
metabolism, it cannot live. When a cell is larger it requires more enzymes and parts
to function correctly, therefore it needs more instruction. In many large cells there
are more than one nuclei to ensure that all cell activities will be carried out quickly.
Surface area-to-volume ratio1: The third reason why cells are limited in size is that
as a cell’s size increases, its volume increases much faster than its surface area.
1mm
Surface Area = 6mm2
Volume = 1 mm3
2mm
Surface Area = 24mm2
Volume = 8 mm3
4mm
Surface Area = 96mm2
Volume = 64 mm3
In the illustration above the smallest cell has a volume of 1mm3 and a surface area of
6mm2. If the side of the cell is doubled to 2 mm, the surface area would increase
fourfold to 24 mm3. When the cell side is 4mm the volume increases to 64 mm3 and
the surface area to 96 mm2.
If cell size is doubled, the cell would require eight times more nutrients and would
have eight times more waste to excrete. The surface area would only increase by a
1
Surface area to volume ratio = surface area / volume
Cell Size Investigation
factor of three. Therefore, the plasma membrane would not to have enough surface
area through which oxygen, nutrients, and waste could diffuse. The cell would either
starve to death or be poisoned by the buildup of waste products.
Objective:
In this lab, the diffusion rate of sodium hydroxide will be observed through 4 agar
cubes of different size. The cubes contain phenolphthalein which turns purple when
it comes in contact with basic solution, sodium hydroxide. Sodium hydroxide will
slowly diffuse into the agar cubes and change its color from clear to purple.
Procedure:
CAUTION!
Sodium hydroxide is caustic! Goggles must be worn when using this
chemical!
1. Obtain pieces of agar. Using a knife cut the agar into cubes with the
measurements below. Plan your cuts ahead of time. Cut out the large cube first.
Block A:
Block B:
Block C:
Block D:
0.5 cm sides
1.0 cm sides
2.0 cm sides
3.0 cm sides
2. Discard remaining agar.
3. Calculate the surface area for each cube and record in table 1.
Surface area = length of a side X width of a side X number of sides
4. Calculate the volume of each cube and record in table 1.
Volume = length X width X height
5. Find and record the surface area to volume ratio for each cube and record in
table 1.
Surface area to volume ratio = surface area / volume
6. Place all four cubes in the bottom of a suitable container so that they do not
touch each other or the sides of the container.
7. Pour in enough 0.1M sodium hydroxide to cover all four cubes.
8. Allow 10 minutes for the diffusion process. GENTLY stir the solution so that
diffusion can take place evenly over each cube.
9. After 10 minutes remove the blocks from the solutions, rinse under water, place
on a paper towel and slice each cube in half.
10. Quickly measure to the nearest 0.1 cm across the area with no sodium hydroxide
on each cube (clear area).
11. Calculate the surface area, volume, and surface area to volume ratio for the
portion of the cube with no sodium hydroxide (clear cube) and record it in table 2.
Clean up:
1. Pour all waste sodium hydroxide down the sink and flush sink with fresh water.
2. Wipe down all working surfaces with damp then dry paper towels.
3. Waste agar may be thrown into trash containers. DO NOT throw waste agar into
sinks.
4. Verify that all sinks are free from any waste.
Cell Size Investigation
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Biology IIH
Cell Size Investigation
Data and Analysis
_________________________
Name
Score _________
Lab Partner __________________________
Data
Table 1:
Cube
Side length (cm)
Surface area (cm2)
Volume (cm3)
Surface area to volume ratio
A
B
C
D
Table 2
Portion of the cube without sodium hydroxide
Cube
Side length of
clear cube (cm)
2
Surface area (cm )
3
Volume (cm )
Volume diffused into the
cube
(original cube volume minus
clear cube volume)
A
B
C
D
Analysis
1. What solution diffused into the agar block? ________________________________
2. Rank the cubes from the largest surface area to volume ratio to the smallest surface
area to volume ratio.
Data and Analysis
3. List and explain three reasons why cells are relatively small in size.
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4. Draw three “cells” (6-sided figures) having approximately the same volume.
Calculate the surface area for each figure. Then, circle the one with the largest
surface area.
A.
B.
C.
5. Construct a one sentence principle that explains the relationship of surface area to
volume as demonstrated in this lab.
___________________________________________________________________
___________________________________________________________________
6. List and explain three ways that cells have evolved to resolve the problems
associated with the relationship of surface area to volume as demonstrated in this
lab
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Data and Analysis
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